Why These Integrals Have a Finite Pattern
For an integral such as \(\int x^4 e^x\,dx\), integration by parts can be applied repeatedly. Each differentiation lowers the degree of the polynomial, while each integration of \(e^x\) leaves \(e^x\) unchanged. Because repeated differentiation eventually turns a polynomial into zero, the process stops after finitely many rows.
A derivative table makes that finite process visible. It also reveals a regular pattern in the coefficients: the powers of \(x\) descend, the signs alternate, and the numerical factors come from successive derivatives of the monomial. The result is an antiderivative that can be checked directly.
How to Read the Table
The left column begins with \(x^n\) and lists its derivatives until the derivative is zero. The right column lists antiderivatives of \(e^x\); every entry is \(e^x\), since \(\int e^x\,dx=e^x\). Multiply entries diagonally, beginning with the first derivative entry and the next antiderivative entry. Give the products alternating signs, starting with \(+\).
For example, the first derivative on the left pairs with the first antiderivative below \(e^x\) on the right, producing \(n x^{n-1}e^x\) with a minus sign. The next derivative pairs with the next antiderivative and receives a plus sign. Continue until the derivative column reaches zero. The final nonzero derivative contributes the last term.
Continue until the next derivative is zero; this ensures the table is finite.
Each antiderivative is \(e^x\), because differentiating \(e^x\) gives \(e^x\).
Start with a positive product, then use negative, positive, negative, and so on.
The terms should cancel in pairs, leaving the original integrand.
Worked Examples
Worked Example: Integrating \(x^2 e^x\)
Differentiate \(x^2\) until the derivative is zero. The antiderivative column repeats \(e^x\):
| Sign | Derivative of \(x^2\) | Antiderivative of \(e^x\) |
|---|---|---|
| \(+\) | \(x^2\) | \(e^x\) |
| \(-\) | \(2x\) | \(e^x\) |
| \(+\) | \(2\) | \(e^x\) |
| \(0\) |
The diagonal products are \(x^2e^x\), \(2xe^x\), and \(2e^x\), with signs \(+\), \(-\), and \(+\). Therefore,
To check, differentiate the proposed antiderivative using the product rule:
The derivative is the original integrand, so the antiderivative is verified. The arbitrary constant \(C\) has derivative zero.
Worked Example: Integrating \(x^3 e^x\)
The successive derivatives of \(x^3\) are \(x^3\), \(3x^2\), \(6x\), \(6\), and then \(0\). Pairing these with the repeated antiderivative \(e^x\) gives:
| Sign | Derivative of \(x^3\) | Antiderivative of \(e^x\) |
|---|---|---|
| \(+\) | \(x^3\) | \(e^x\) |
| \(-\) | \(3x^2\) | \(e^x\) |
| \(+\) | \(6x\) | \(e^x\) |
| \(-\) | \(6\) | \(e^x\) |
| \(0\) |
Thus the signed products give
For a check, set \(P(x)=x^3-3x^2+6x-6\). Then \(P'(x)=3x^2-6x+6\), so
The polynomial terms of degrees two, one, and zero cancel, leaving exactly the cubic term.
Worked Example: Integrating \(x^4 e^x\)
Here the derivatives of \(x^4\) are \(x^4\), \(4x^3\), \(12x^2\), \(24x\), \(24\), and then \(0\). The table therefore has five nonzero diagonal products:
| Sign | Derivative of \(x^4\) | Antiderivative of \(e^x\) |
|---|---|---|
| \(+\) | \(x^4\) | \(e^x\) |
| \(-\) | \(4x^3\) | \(e^x\) |
| \(+\) | \(12x^2\) | \(e^x\) |
| \(-\) | \(24x\) | \(e^x\) |
| \(+\) | \(24\) | \(e^x\) |
| \(0\) |
The signs and products produce
Check the result by writing \(P(x)=x^4-4x^3+12x^2-24x+24\). Its derivative is \(P'(x)=4x^3-12x^2+24x-24\). Hence
Every term below degree four cancels with its counterpart from \(P'(x)\), confirming the answer.
The General Coefficient Pattern
The examples show the same structure at each degree. The leading coefficient is \(1\). The next coefficient is the negative of the degree. The following coefficient is the degree multiplied by the next lower integer, and the signs continue to alternate. For a nonnegative integer \(n\), the complete result is
More precisely, the polynomial in parentheses is
This expression makes the coefficient rule precise. At \(k=0\), the coefficient is \(1\). At \(k=1\), it is \(-n\). At \(k=2\), it is \(n(n-1)\). The final term occurs at \(k=n\): its power is \(x^0=1\), and its coefficient is \((-1)^n n!\). The factorial \(m!\) means \(1\cdot2\cdots m\) for positive integers \(m\), with \(0!=1\).
Why the Formula Works
The table is an organized repetition of integration by parts. Its signs alternate because each use of integration by parts subtracts the next diagonal product. The process ends because the derivative of a polynomial of degree \(n\) becomes zero after \(n+1\) differentiations. The cancellation can also be proved directly by differentiating the proposed form.
Let \(P_n(x)\) be the polynomial above. Its coefficient on \(x^r\), for \(0\le r\le n\), is
For \(0\le r<n\), the coefficient on \(x^r\) in \(P_n'(x)\) comes from differentiating the \(x^{r+1}\) term. That coefficient is
This is the negative of the coefficient on \(x^r\) in \(P_n(x)\). Thus every power below \(x^n\) cancels in \(P_n+P_n'\). The \(x^n\) term has coefficient \(1\) in \(P_n\) and has no contribution from \(P_n'\), so
Using the product rule now proves the formula:
Therefore \(e^xP_n(x)+C\) is an antiderivative of \(x^n e^x\). This proof also explains the table's alternating pattern: each polynomial term is canceled by the derivative of the adjacent term, except for the original leading power.
A Common Sign Error
Alternating signs are essential. A common mistake is to make every term positive. For instance, the incorrect proposal
does not integrate \(x^2e^x\). Differentiating it gives
which is not \(x^2e^x\). The table supplies a minus sign for the \(2x\) diagonal product, so the corrected polynomial is \(x^2-2x+2\). Differentiating that expression gives \(x^2e^x\), as the first worked example confirms.
When the Table Applies
This finite table is especially direct when the algebraic factor is a polynomial and the other factor is \(e^x\). It relies on two facts: differentiating the polynomial eventually gives zero, and every successive antiderivative of \(e^x\) is still \(e^x\). The degree \(n\) must be a nonnegative integer for the derivative column to terminate in this way.
For \(n=0\), there is just one nonzero row, and the formula gives \(\int e^x\,dx=e^x+C\). For larger \(n\), the new terms follow the same descending-power, alternating-sign pattern. Always include \(+C\) for an indefinite integral, and differentiate the complete proposed antiderivative when checking it.
Check Your Understanding
Use the finite derivative table and the cancellation in the product rule to answer these questions.
- What are the successive nonzero derivatives of \(x^3\), and what signs are assigned to their diagonal products?
- Write the coefficient pattern in the antiderivative of \(x^4e^x\), including its constant term.
- Why does the derivative column in the table eventually become zero?
- Differentiate \(e^x(x^3-3x^2+6x-6)\) and show which terms cancel.
- What goes wrong if the second diagonal product is given a positive sign instead of a negative sign?
